Quantitative Aptitude - Arithmetic Ability Questions

$$\begin{gathered} 112 × 54 \\ 112 × 1024 \\ = 112 × 1000024 \\ = 1120000/16 \end{gathered}$$

= 70,000.

Given,
S.I = 3 Principal Amount
=> 3A = A x 16 x R/100
By solving, we get
=> R = 18.75%

Difference in amount = 514 - 415 = 99

99 = (415 x 4 x R)/100

R = 5.96%

1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14

Sum of 4 digit numbers = (2+4+6+8) x P33 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x P23 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x P13 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x P03 x (1) = 20 x 1 x 1 = 20 

Adding All , Sum = 147320

Let the cost price of the fruits be 'C.P'

From the given data, after analysis it can be solved as

(121% of C.P) - (125% of 91% of C.P) = 29

$$\begin{gathered} 121100 C.P - 125100 x 91100 x C.P = 29 \\ 121100 - 125 x 91100 x 100 C.P = 29 \\ (12100 - 125 x 91) C.P = 290000 \\ (12100 - 11375) C.P = 290000 \\ 725 C.P = 290000 \\ C.P = 400 \end{gathered}$$

Hence, the cost price of the fruits = Rs. 400.

sum = [(S.I x T.D)/(S.I-T.D)] = Rs. [(375 x 250)/(375-250)] = Rs.750 

Rate = (100 x 375)/(750 x 3) =1623 %

The total possible cases would be a 5 card hand with no restrictions :52C5 5

The unwanted cases are:

no queens(out of 48 non-queens cards we get 5) 48C5

only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4) 4C1*48C4

Therefore,52C5-(48C5+4C1*48C4) = 108336